Question

Find the area ratio of a regular octahedron and a tetrahedron regular, knowing that the diagonal of the octahedron is equal to height of the tetrahedron.

Answer 1

Answer:

$\frac{4}{3}$

Step-by-step explanation:

The area of a regular octahedron is given by:

area = $2\sqrt{3}\ *edge^2$. Let a is the length of the edge (diagonal).

area = $2\sqrt{3}\ *a^2$

Given that the diagonal of the octahedron is equal to height (h) of the tetrahedron i.e.

a = h, where h is the height of the tetrahedron and a is the diagonal of the octahedron. Let the edge of the tetrrahedron be e. To find the edge of the tetrahedron, we use:

$h=\sqrt{\frac{2}{3} } e\\but\ h=a\\a=\sqrt{\frac{2}{3} } e\\e=\sqrt{\frac{3}{2} }a$

The area of a tetrahedron is given by:

area = $\sqrt{3}\ *edge^2$ = $\sqrt{3} *(\sqrt{\frac{3}{2} }a)^2=\frac{3}{2}\sqrt{3} *a^2$

The ratio of area of regular octahedron to area tetrahedron regular is given as:

Ratio = $\frac{2\sqrt{3}\ *a^2}{\frac{3}{2} \sqrt{3}*a^2} =\frac{4}{3}$